The harmonic oscillator with damping
Definition:
Summary:
• body of mass m attached to spring with spring
constant k is released from position x0 (measured
from equilibrium position) with velocity v0;
The equation of motion is
d 2 x(t)
dx(t)
+ 2β
+ ω20 x(t) = 0
2
dt
,
dt
• resistance due to friction F res = − bv , b = nonnegative constant (possibly zero)
x(t)
where
• β=
b
2m
and
ω0 =
k
.
m
There are three distinct kinds of motion:
Prerequisites:
• β<ω0 : underdamped
• fundamentals of Newtonian mechanics
• β=ω0 : critically damped
• t h e simple harmonic oscillator without resistance
• β>ω0 : overdamped
Go to derivation.
Why study it?
• a very simple dynamical system with an exact
solution in closed form;
• occurs frequently in everyday applications
Go to Java™ applet
In an earlier section, we studied the simple harmonic oscillator. Our example was an ideal
spring with no damping force - no friction
between the body and the track. Now let’s see
what happens when we relax this restriction, and
include a resistive force of the form
Fres = –bv ,
where b is a positive constant and v is the
velocity. (We considered such a force in another
context in a previous section.)
First, we try to picture what will happen. The
resistive force will act to slow the motion down,
and the amplitude of the oscillation will
continually decrease. We might expect this
decrease to be exponential, based on our earlier
experience with this form of resistive force.
We might also expect there to be several cases,
depending on the amount of damping. Imagine a
screen door which is released from its open
position. If the damper is worn out and doesn’t
provide enough damping, the door will slam shut.
(If the door frame were not there, the door would
oscillate back and forth a few times before
stopping.)
If the damper is really stiff, the door might take
too long to shut, letting in all kinds of bugs.
Somewhere in between these two cases will be a
case where the damping is just the right amount
to allow the door to close in the least amount of
time without slamming. Here is a movie illustrating these three cases.
Now let’s see how well our intuition holds up.
Adding up the forces on the body and using
Newton’s second law, we find
ma(t) = − kx(t) − bv(t) ,
where the first term on the right-hand side is the
restoring force due to the spring, and the second
is the damping force due to friction. Rearranging,
we find the differential equation
d 2 x(t)
b dx(t)
+
+ ω20 x(t) = 0 ,
2
m dt
dt
where ω0 = k / m is the natural freqency of the
oscillator.
It takes some inspiration to solve this equation.
The standard trick is to try a solution of the form
exp(αt) ,
because this function just reproduces itself when
differentiated. The value of the constant α is
determined by plugging this form back into the
differential equation. We find
(α2 +
b
α + ω 20 ) exp(α t) = 0 ,
m
which can only be true for all t if the first factor
is zero. Using the quadratic equation, we find
äåå b 2
ë
b
± åå 2 – ω20 ìììì
α=–
2m ã 4m
í
1/2
.
We will make the following shorthand notation:
b
β≡
.
2m
There will be three cases, depending on the size
of β (the amount of damping).
This is the case of small damping. The
argument of the square root in α is negative, so α
can have either of the two complex values
α = –β ± i ω20 – β2 ,
where i2 =–1. (If you are weak on complex
numbers, now would be the time to review.) As a
shorthand, we will write
ω1 =
ω –β
ã
βx 0 + v 0
ìë
sin ω1 t ìììì
ω1
í
.
A second case occurs when
2) β=ω0 (“critically damped”)
In this case, there is only one value of α:
α = –ω0 .
The solution is the limit of the underdamped case
as ω1 goes to zero:
1) β<ω0 (“underdamped”)
2
0
åä
x(t) = exp á − βt é åååå x 0 cos ω1 t +
2
.
One of the basic results of complex analysis is
that exp(iω 1 ) = cos ω1 + i sin ω1. Hence, our
solutions are linear combinations of the functions
exp á –βt é cos ω1 t and exp á –βt é sin ω1 t .
The solution is easily found to be
x(t) = exp á –βt é á x 0 + á βx 0 + v 0 é t é
The linear dependence on t is characteristic of the
case in which the two possible values of α are
equal.
3) β>ω0 (“overdamped”)
In this case, there are again two values of α, this
time both real:
α = –β ±
β2 – ω20 .
As a shorthand, we will write
ω2 =
β 2 – ω20
.
size of β:
The solution is
äå
x(t) = exp á − βt é åååå x 0 cosh ω 2 t +
ã
ëì
βx 0 + v 0
sinh ω2 t ìììì
ω2
í
t
where cosh and sinh are hyperbolic functions.
The following plot shows typical curves for each
of the three cases, all with the same x0 and with
v0=0:
β
Here is a movie illustrating the three kinds of
damping.
Damping:
critical
over
under
under
2. Damped, driven oscillator
t
We see that the motion is a decaying oscillation
in the underdamped case; the amplitude decays
according to the envelope exp(–βt). We see that
equilibrium is approached fastest for the
critically damped case, hence its name.
Here is a three-dimensional plot showing how the
three cases go into one another depending on the
You may recall our earlier treatment of the driven harmonic oscillator with no damping. We
found that if the driving frequency is equal to the
natural frequency of the system, then the
amplitude becomes arbitrarily large as time goes
on.
This is unphysical, however. In practice,there is
always some damping present. As the velocity
becomes larger, this damping leads to greater and
greater energy loss from the system, as we
discussed earlier. As time goes on, the rate of
energy lost due to damping balances the energy
gained due to the external driving force, and a
steady-state oscillation is achieved.
Let’s see how this is reflected in the
mathematics. Newton’s law now reads
d 2 x(t)
dx(t)
m
+b
+ k x(t) = F 0 cos (ω t) .
2
dt
dt
The solution is quite messy, but otherwise
straightforward. It can be arranged into the sum
of two pieces, one of which is proportional to
exp(–βt) and the other of which is not. The
former goes away as time becomes large, and is
therefore called a transient. We will ignore this
piece.
The other piece, the steady-state solution, is the
one we are interested in. It is easy to show by
direct substitution into the differential equation
that it is given by
xs(t) =
F0
m (ω2 − ω2R ) 2 + 4β2 ω21
cos(ωt − δ)
where the resonant frequency is given by
ωR =
and the phase satisfies
ω20 − 2β2
,
tan δ =
2βω
ω20 − ω2
.
(The quantity ω21 = ω20 − β2 was defined earlier.)
These are very important equations, and it is
worth spending some time studying their
properties (rather than deriving them).
The amplitude of the steady-state solution is
F0
A(ω) =
.
m (ω2 − ω2R )2 + 4β2 ω21
For the moment, let’s regard the damping β as
fixed and the driving frequency ω as variable.
The denominator of A(ω) is smallest when the
driving frequency is equal to the resonant
frequency. So A(ω) itself is a maximum for that
value of ω. This is called resonance. (In the last
section, we found resonance at ω = ω0 when no
damping is present. This agrees with our present
formula for the resonant frequency, when β=0.)
Looking at our formula for the resonant
frequency, we see that the resonance effect only
occurs when ω0 > 2 β, and we restrict ourselves
to this case.
If we plot the amplitude of the steady-state
solution versus ω, we get a curve with a peak at
ω0
decreasing
damping β
Now let’s plot δ, the phase by which the steadystate solution lags behind the driving force.
δ
01
0101
0101
0101
0101
0101
0101
0101
1001
A(ω)
0101010101010101010101
the resonant frequency. The next diagram shows
several such curves, each for a different value of
the damping constant β.
π
π/2
ω
The height and width of the peak are controlled
mainly by the value of β. The height of the peak
is
F0
,
2mω 1 β
and its width is proportional to
2ω 1 β .
The smaller the damping β, the higher and
narrower the peak gets, and the closer the
resonant frequency gets to the natural frequency
ω0.
ω0
ω
(The curves have the same color as their partners
in the previous plot.)
We see that there is always a delay between the
action of the driving force and the response of the
system. When the system is being forced at its
natural frequency ω0 , the phase lag is π/2. You
might be familiar with this if you have ever
played on a swingset. You can increase your
amplitude the fastest by leaning back right at the
bottom of your swing, one-quarter cycle behind
your maximum amplitude.
If the driving frequency is much smaller than the
natural frequency, the driving force and the
response are in phase. This makes sense,
because the system can adjust itself to match the
slow driving force. On the other hand, if the
driving frequency is much higher than the natural
frequency, the driving force and the response are
out of phase (phase lag π).